Monday, March 22, 2021

Molecular Basis of Inheritance | Class 12 | Sample Questions and Answers

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MOLECULAR BASIS OF INHERITANCE


QUESTIONS


1 Score Questions

1.   Which of the following combinations do not apply to DNA?

(a) Deoxyribose, Guanine               

(b) Ribose, Adenine

(c) Deoxyribose, Uracil                    

(d) Guanine, Thymine


(1)   (a) and (b)                     

(2)   (b) and (c)

(3)   (c) and (d)                     

(4)   (a) and (d)  

2.   In DNA, the proportion of A is equal to T and the proportion of G is equal to C. This rule is known as ………

3.   Biochemical characterization experiment proved that transforming principle is

        (a) RNA                                     

        (b)  DNA

        (c) Protein                               

        (d)  Lipid

4.   During DNA replication, one strand is synthesised in small stretches called …………

5.   Note the relationship between first two words and fill up the fourth place.

UGG: Tryptophan                 

AUG: …………………

6.   Expand the abbreviation VNTR.

2 Score Questions

7.   In a DNA segment, cytosine is 22%. Find out the percentage of adenine, thymine and guanine.

8.   Identify the structure given below. Label A, B and C.

9.   Nucleosomes constitute the repeating unit to form chromatin.

a.   Name the two forms of chromatin.

b.   Mention any two differences.

10.  Following are the first two steps in Griffith’s transformation experiment.

               i.  S-strain → inject into mice → mice live

             ii.   R-strain → inject into mice → mice die

a.   If there is any mistake correct it.

b.   Write the remaining steps.

11.  Blender experiment (Hershey and Chase experiment) proved that DNA is the genetic material.

a.     Name the organisms used for this experiment.

b.    What are the steps involved in this experiment?

12.  DNA is the better genetic material than RNA. Do you agree with this statement? Substantiate.

13.  The flow of genetic information is shown below:

 Name the processes a, b, c and d.

14.  Results of a famous experiment is given in the figure.

a.   Identify the experiment.

b.   Which property of DNA is proved by experiment?

15. The diagram of an mRNA is given below. Copy it and mark the 5’ end 3’ end of mRNA by giving reason.

16. Read the sequence of codon in the mRNA unit.

a.   What changes is needed in the first codon to start the translation process?

b.   If translation starts by that change, till which codon it can be continues?

17. Observe the diagram and answer the question.

a.   Identify the process and its site.

b.   What is the role of mRNA and tRNA in this process?

18. Human Genome Project (HGP) was the first mega project for the sequencing of nucleotides and mapping of all the genes in human genome.

a.  What are the two main approaches in the methodologies of HGP?

b.  Expand BAC & YAC.

19. Schematic representation of DNA finger prints is given.

a.  Which one of the suspected individuals may involve in the crime?

b.  Write any other use of DNA fingerprint.

3 Score Questions

20. Observe the diagram and answer the question.

a.  What is meant by replication fork?

b.  What is the difference in the replication process in strand A and strand B?

c.  What is the role of DNA ligase in the replication process in B strand?

21. A transcription unit is given below. Observe it and answer the question.

    a. How can you identify the coding strand?

    b. Write the sequence of RNA formed from this unit.

    c. What would happened if both strand of DNA act as template for transcription?

22.  Given below is the diagrammatic representation of first stage of a process in a bacterium.

a.   Identify the process.

b.   Name the enzyme catalyses this process.

c.   What are the additional complexities in eukaryotes in this process?

23.  With the help of the figure given, explain the processing of hnRNA to mRNA in eukaryotes.

24.  Observe the figure of mRNA.

a.  Find the start codon and stop codon.

b.  How many amino acids will be present in the protein translated from this mRNA?

c.  The additional sequences that are not translated in the mRNA is called ................

25. “Prediction of the sequence of amino acids from the nucleotide sequence in mRNA is very easy, but the exact prediction of nucleotide sequence in mRNA from the sequence of amino acids coded by mRNA is difficult”

a.  Which property of genetic code is the reason for the above condition? Explain.

b.  Which are the stop codons in DNA translation?

26.  A DNA sequence for coding a peptide is given below:

3’-CAAGTAAATTGAGGACTC-5’


Codons

Amino acids

UUA

Leu

CCU

Pro

CAU

His

ACU

Thr

GUU

Val

GAG

Glu

a.  Write the sequence of complementary mRNA.

b.  Find out the amino acids sequence of peptide chain using the codon given in the hints.

c.  Coding region of a gene consists of 450 nucleotide base pairs. Avoiding stop codons & introns, how many amino acids would the corresponding polypeptide chain contain? Justify your answer.

27.  In an E. coli culture, lactose is used as food instead of glucose.

a.   How do the bacteria respond to this situation at genetic level?

b.   If lactose is removed what will happen?

28. Lac operon in the absence of inducer (Lactose) is given.

a.  What is ‘P’?

b.  Name the enzyme produced by the structural gene ‘Z’,’Y’, and ‘A’.

c.  Redraw the diagram in the presence of an Inducer.

29. a) The steps in DNA Finger printing are given below. Complete the flow chart (A and B).

b) Mention the application of DNA finger printing.

30. A small fragment of a skin of different person was extracted from nails of a murdered person. This led the crime investigators to the murder.

a.  What technique was used by the investigators?

b.  What is the procedure involved in this technique?


ANSWERS

1.   2. (b) and (c)

2.   Erwin Chargaff’s rule.

3.   (b) DNA

4.   Okazaki fragments.

5.   Methionine.

6.   Variable Number Tandem Repeats.

7.   According to Erwin Chargaff’s rule, in DNA, proportion of A is equal to T and the proportion of G is equal to C.

If cytosine (C) = 22%, then guanine (G)= 22%. (C+G = 44%.)

So A+T = 100 – 44 = 56%.

Adenine (A) = 28% and Thymine (T) = 28%.

8.   Nucleosome. A= DNA, B= H1 histone, C= Histone octamer.

9.   (a) Euchromatin and Heterochromatin.

(b) Euchromatin: Loosely packed and transcriptionally active region of chromatin. It stains light.

Heterochromatin: Densely packed and inactive region of chromatin. It stains dark.

10. (a) i. S-strain → inject into mice → mice die

ii. R-strain → inject into mice → mice live

(b) S-strain (Heat killed) → Inject into mice → Mice live

S-strain (Hk) + R-strain (live) → Inject into mice → Mice die.

11. (a) E. coli & Bacteriophage.

(b) Infection bacteriophage with E. coli.

Blending to remove the virus particles from the bacteria.

Centrifugation to separate virus particles from bacterial cells.

12. Yes. For the storage of genetic information, DNA is better due to its stability.

13. (a) Replication. (b) Transcription. (c) Translation

(d) Reverse transcription.

14. (a) Meselson & Stahl experiment.

(b) Semi-conservative model of DNA replication.

15. Here, one end of mRNA has polyadenylate residues (AAAAA...). Polyadenylation (tailing) occurs at 5’ end.

16. (a) First codon should be AUG.

(b) Translation continues till 5th codon (UUU) because after that the stops codon UGA comes. So translation is stopped.

Read carefully the sequence of codon in the mRNA unit and answer the question.

17. (a) Translation (Protein synthesis). Its site is ribosome.

(b) mRNA: Provide template for translation.

tRNA: Brings amino acids for protein synthesis and reads the genetic code.

18. (a) Expressed Sequence Tags (ESTs) & Sequence annotation.

(b) BAC= Bacterial Artificial Chromosomes.

YAC= Yeast Artificial Chromosomes.

19. (a) Individual B.

(b) For the diagnosis of genetic diseases.

20. (a) Unwinding of the DNA molecule at a point forms a ‘Y’-shaped structure called replication fork.

(b) In strand A, continuous synthesis occurs. In strand B, discontinuous synthesis occurs.

(c) It helps to join the Okazaki fragments formed (due to discontinuous synthesis) together to form a new strand.

21. (a) Coding strand will be in 5’-3’ direction.

(b) 5’ – U C A G U A C A – 3’

(c) If 2 RNA molecules are produced simultaneously, this would be complimentary to each other. It forms a double stranded RNA and prevents translation

22. (a) Transcription.

(b) DNA-dependent RNA polymerase.

(c) There are 3 RNA polymerases (RNA polymerase I, II & III). The primary transcripts (hnRNA) contain exons and introns and are non-functional. Hence introns must be removed.

23. Steps of processing of hnRNA to mRNA:

Splicing: From hnRNA, introns are removed (by the spliceosome) and exons are spliced (joined) together.

Capping: Here, a nucleotide methyl guanosine triphosphate (cap) is added to the 5’ end of hnRNA.

Tailing (Polyadenylation): Here, adenylate residues (200-300) are added at 3’-end.

24. (a) Start codon is AUG (3rd codon). Stop codon is UAG.

(b) 4 amino acids.

(c) Untranslated Regions (UTRs).

25. (a) Degeneracy. An amino acid (except methionine & tryptophan) is coded by more than one codon. They are called degenerate codons.

(b) UAA, UAG & UGA.

26. (a) 5’-GUU CAU UUA ACU CCU GAG-3’

(b) Val – His – Leu – Thr – Pro – Glu

(c) 150 amino acids. A codon contains 3 bases. So in this gene, 150 codons are there. A codon codes for one amino acid. Thus 150 codons code for 150 amino acids.

27. (a) If lactose is provided in the growth medium, it is transported into E. coli cells by the action of permease. Lactose (inducer) binds with repressor protein. So repressor protein cannot bind to operator region. The operator region becomes free and induces the RNA polymerase to bind with promoter. Then transcription starts.

(b) If there is no lactose, lac operon remains switched off. The regulator gene synthesizes mRNA to produce repressor protein. This protein binds to the operator region and blocks RNA polymerase movement. So the structural genes are not expressed.

28. (a) P is the promoter for structural genes.

(b) z gene: b galactosidase, y gene: Permease

a gene: Transacetylase.

(c)  

29. (a) A = Digestion of DNA by restriction endonucleases. B= Hybridization using radioactive labelled VNTR probe.

(b) Forensic tool to solve paternity, rape, murder etc.

For the diagnosis of genetic diseases.

To determine phylogenetic status of animals.

To determine population and genetic diversities.

30. (a) DNA fingerprinting.

(b) Procedures:

               i.   Isolation of DNA.

               ii.  Digestion of DNA by restriction endonucleases.

               iii. Separation of DNA fragments by gel electrophoresis.

               iv. Transferring DNA fragments to synthetic membranes such as nitrocellulose.

               v.  Hybridization using radioactive labelled VNTR probe.

               vi. Detection of hybridized DNA by autoradiography.

Chapter-wise Sample Q & A

CLASS 11 (PLUS 1) BOTANY: QUESTIONS & ANSWERS

CLASS 11 (PLUS 1) ZOOLOGY: QUESTIONS & ANSWERS

CLASS 12 (PLUS 2) BOTANY: QUESTIONS & ANSWERS

CLASS 12 (PLUS 2) ZOOLOGY: QUESTIONS & ANSWERS

Chapter-wise Previous Year Q & A

CLASS 11 (PLUS 1) BOTANY: CHAPTER-WISE Q & A

CLASS 11 (PLUS 1) ZOOLOGY: CHAPTER-WISE Q & A

CLASS 12 (PLUS 2) BOTANY: CHAPTER-WISE Q & A

CLASS 12 (PLUS 2) ZOOLOGY: CHAPTER-WISE Q & A

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